Ex 11.3, 6 - Find equation of planes that passes through 3 points

Ex 11.3, 6 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

Ex 11.3, 6 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Ex 11.3, 6 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

 

 

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Question 6 (Introduction) Find the equations of the planes that passes through three points. (a) (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3) Vector equation of a plane passing through three points with position vectors 𝑎 ⃗, 𝑏 ⃗, 𝑐 ⃗ is ("r" ⃗ − 𝑎 ⃗) . [(𝑏 ⃗−𝑎 ⃗)×(𝑐 ⃗−𝑎 ⃗)] = 0 Question 6 Find the equations of the planes that passes through three points. (a) (1, 1, –1), (6, 4, –5), (–4, –2, 3) Vector equation of a plane passing through three points with position vectors 𝑎 ⃗, 𝑏 ⃗, 𝑐 ⃗ is ("r" ⃗ − 𝒂 ⃗) . [(𝒃 ⃗−𝒂 ⃗)×(𝒄 ⃗−𝒂 ⃗)] = 0 Now, the plane passes through the points (𝒃 ⃗ − 𝒂 ⃗) = (6𝑖 ̂ + 4𝑗 ̂ – 5𝑘 ̂) − (1𝑖 ̂ + 1𝑗 ̂ − 1𝑘 ̂) = (6 −1)𝑖 ̂ + (4 − 1)𝑗 ̂ + (−5 − (−1)) 𝑘 ̂ = 5𝒊 ̂ + 3𝒋 ̂ − 4𝒌 ̂ A (1, 1, −1) 𝑎 ⃗ = 1𝑖 ̂ + 1𝑗 ̂ − 1𝑘 ̂ B (6, 4, −5) 𝑏 ⃗ = 6𝑖 ̂ + 4𝑗 ̂ − 5𝑘 ̂ C ( −4, −2, 3) 𝑐 ⃗ = −4𝑖 ̂ − 2𝑗 ̂ + 3𝑘 ̂ (𝒄 ⃗ − 𝒂 ⃗) = (−4𝑖 ̂ − 2𝑗 ̂ + 3𝑘 ̂) − (1𝑖 ̂ + 1𝑗 ̂ − 1𝑘 ̂) = (−4 − 1)𝑖 ̂ +(−2 − 1)𝑗 ̂ + (3 − (−1)) 𝑘 ̂ = −5𝒊 ̂ − 3𝒋 ̂ + 4𝒌 ̂ (𝒃 ⃗ − 𝒂 ⃗) × (𝒄 ⃗ − 𝒂 ⃗) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@5&3&−4@−5&−3&4)| = – |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@5&3&−4@ 5& 3&−4)| = 𝟎 ⃗ This implies, the three points are collinear. Using property: Since the two rows of the determinant are same, the value of determinant is zero. ∴ Vector equation of plane is [𝑟 ⃗−(𝑖 ̂+𝑗 ̂ −𝑘 ̂ )] . 0 ⃗ = 0 Since, the above equation is satisfied for all values of 𝑟 ⃗, Therefore, there will be infinite planes passing through the given 3 collinear points.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo